Piston compression A piston is seated at the top of a cylindrical chamber with radius 5 cm when it starts moving into the chamber at a constant speed of 3 cm/s (see figure). What is the rate of change of the volume of the cylinder when the piston is 2 cm from the base of the chamber? <IMAGE>

Verified step by step guidance

1. First, identify the formula for the volume of a cylinder, which is given by: <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>V</mi><mo>=</mo><mi>&#960;</mi><msup><mi>r</mi><mn>2</mn></msup><mi>h</mi></mrow></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>r</mi></math> is the radius and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi></math> is the height.
2. Since the radius is constant at 5 cm, the formula simplifies to: <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>V</mi><mo>=</mo><mi>&#960;</mi><mo>&#xD7;</mo><mn>25</mn><mo>&#xD7;</mo><mi>h</mi></mrow></math>.
3. To find the rate of change of the volume, differentiate the volume formula with respect to time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi></math>. This gives: <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mfrac><d><mi>dV</mi></d><d><mi>dt</mi></d></mfrac><mo>=</mo><mi>&#960;</mi><mo>&#xD7;</mo><mn>25</mn><mo>&#xD7;</mo><mfrac><d><mi>dh</mi></d><d><mi>dt</mi></d></mfrac></mrow></math>.
4. The problem states that the piston moves at a constant speed of 3 cm/s, which means <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><d><mi>dh</mi></d><d><mi>dt</mi></d></mfrac></math> is 3 cm/s.
5. Substitute <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><d><mi>dh</mi></d><d><mi>dt</mi></d></mfrac></math> = 3 cm/s into the differentiated equation to find the rate of change of the volume: <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mfrac><d><mi>dV</mi></d><d><mi>dt</mi></d></mfrac><mo>=</mo><mi>&#960;</mi><mo>&#xD7;</mo><mn>25</mn><mo>&#xD7;</mo><mn>3</mn></mrow></math>.

Key Concepts

Volume of a Cylinder

The volume of a cylinder is calculated using the formula V = πr²h, where r is the radius and h is the height. In this scenario, the radius is constant at 5 cm, while the height changes as the piston moves. Understanding this formula is essential to determine how the volume changes as the piston compresses the air in the chamber.

Related Rates

Related rates involve finding the rate at which one quantity changes in relation to another. In this problem, we need to relate the rate of change of the volume of the cylinder to the rate at which the height of the piston changes. By applying the chain rule, we can express the rate of change of volume in terms of the rate of change of height.

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate of change of a function. In this context, we will differentiate the volume formula with respect to time to find the rate of change of volume as the piston moves. This process allows us to calculate how quickly the volume of the cylinder is decreasing as the piston compresses the air.